Q186 · Practice questionComplex unfamiliar7 marks
QUESTION 186 (7 marks)
A hydrogen atom emits a photon as its electron falls from to . Use . A second hydrogen photon has wavelength and also begins from .
a)[3 marks]
Calculate the wavelength of the first photon.
b)[3 marks]
Determine the final principal quantum number for the second photon.
c)[1 mark]
Identify which of these two transitions releases more energy, giving a reason.
WORKED SOLUTION
7 marksPractice marking scheme
ANSWER
(a) (). (b) . (c) The transition; its photon has the shorter wavelength.
Worked solution
(a) . Hence .
(b) . Thus , identifying level 3.
(c) , so the shorter wavelength of the transition corresponds to greater emitted energy.
Equivalent physically justified methods and consistent equivalent units accepted.
Displayed decimals are model answers; accept appropriate significant figures and consistent rounding from stated constants.
Part a: Use the final-minus-initial inverse-square level expression.
Part a: Calculate the reciprocal wavelength.
Part a: Invert to obtain wavelength in length units.
Part b: Rearrange the Rydberg relation for the final level.
Part b: Calculate its inverse-square value.
Part b: Identify the consistent integer quantum number.
Part c: Identify the first transition and justify with inverse energy–wavelength dependence.
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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